today we're going to talk about how to use the reduction formula to evaluate an integral
and in this particular problem we've been given the integral of the natural log or ln
of x raised to the third power and I've written the reduction formula here for us for reference
basically a reduction formula is a formula that tells you how to evaluate the integral
so in this particular case it says if you're given the integral of the quantity of ln of
x raised to the n and in this case n is a constant and we have r ln of x raised to a
constant of three so if you're given an integral in exactly this form this is the evaluation
of the integral it's a reduction formula and so all we need to do instead of trying to
figure out how to solve this integral we can just use a formula that we know in advance
and plug in our values so we just need to know that in our case n is three and we plug
three in for n in our formula and we've now reduced the integral so let's take a look
at how this is going to work keep in mind that as we said r value of n is three because
natural log of x is raised to the third power here so we're going to be plugging three in
for n wherever we see n over here on the right hand side and x of course will just stay as
it is so the reduction of our original integral the quantity ln of x to the third power will
be equal to we'll start here with this x it will be equal to x times the natural log of
x raised to the n and in our case n is three so raised to the third minus again here we
have n and our n is three so three times the integral of the natural log of x to the n
minus one power well in our case since n is three we'll get three minus one and we'll
have two here so we have ln of x squared dx now this is why it's called a reduction formula
notice that our original integral was the quantity natural log of x cubed our new integral
is the quantity natural log of x squared so we have reduced it by a power or by a degree
the degree has gone down from three to two so we have a reduction of the degree of the
original function that we we're taking the integral of so we still haven't solved completely
for the integral we have this remaining integral quantity ln of x squared but we can apply
the reduction formula again because now we have the same form as our original integral
but our new n value is just two so we can apply this reduction formula again with n
equal to two so what we'd have in that case this front part here would stay x ln of x
cubed minus three now the right hand side of our reduction formula here is what we put
in place of the integral so everything in front of this integral here up to the three
stays and we can draw a big parenthesis and now in place of this remaining integral we
plug in the right hand side so again we start with x and we have x times the quantity of
ln of x raised to the n power and in this case our new value for n is two so two minus
here we have n in front of the integral so again we have two here times the integral
of the quantity ln of x raised to the n minus one so that's for us two minus one which is
just one and we don't have write that but we would have one here and then just dx and
then we can close our bracket or our parenthesis and again you see we've now reduced the degree
of our function from two down to one when we had ln of x cubed then we have ln of x
squared and now we have ln of x to the first power so we still have an integral we need
to keep going we can just use the reduction formula again this time with the value of
n as one because the degree of our ln of x here is just one so before we do that let's
go ahead and simplify so we can get rid of these brackets and we don't get confused later
we'll just have x ln of x cubed minus we're going to just distribute this negative three
across the two terms we have inside here so we'll get minus three x times ln of x squared
then we have a negative three times a negative two which is going to give us a positive six
times the integral of ln of x dx and remember we just have this is essentially ln of x to
the first power so now we'll use our reduction formula again to evaluate this integral here
so we'll keep everything in the front and our reduction formula comes in place of our
remaining integral so again here we start with x the right hand side of our reduction
formula here x times ln of x raised to the n power in our case n is one and now we just
have that one minus here we have another one we don't need to write it but we'd have one
here so one times the integral of ln of x now we have one minus one right our new n
is one so we get one minus one which is just zero so we have we have ln of x to the zero
power and then dx now this is interesting because ln of x raised to the zero is just
one anything raised to the zero power is just one so we can go ahead and simplify this integral
it would be now one times dx so let's re-write it and let's go ahead and distribute the six
we'll get six x times ln of x we'll have six and then minus the integral so we'll get minus
six times the integral remember ln of x to the zero anything raised to the zero power
is one so we get one times dx so essentially we just have the integral of dx and that's
it well the integral of dx if we take the integral of dx we just get x so the integral
of dx is x which means that our final answer is just going to be
everything in front minus a six x and then don't forget because we're taking an integral
here to add c to account for the constant of integration but this is our final answer
and that's how we use a reduction formula and how we can use it over and over again
to continue simplifying an integral like this until we arrive at our final answer oh and
keep in mind that if you want to here you can factor out an x and get x times ln of
x to the third minus three times ln of x squared plus six times ln of x minus six and then
plus c factoring out that x will simplify this final answer even further so I hope you
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